Transcript AC Series

Lesson 20
Series AC Circuits
Learning Objectives

Compute the total impedance for a series AC
circuit.

Apply Ohm’s Law, Kirchhoff’s Voltage Law
and the voltage divider rule to AC series
circuits.

Graph impedances, voltages and current as
a function of phase.

Graph voltages and current as a function of
time.
Review
Impedance

Impedance is a complex quantity that can be
made up of resistance (real part) and
reactance (imaginary part).
Z = R + jX = Z

Unit of impedance is ohms ().
()
Review
Resistance

For resistors, voltage and current are in phase.
Z R  R  0 j  R0
Review

For inductors, voltage leads current by 90º.
Z L  0  j L   L90  X L90  X L j
Review

For capacitors, voltage lags current by 90º.
 1   1 
ZC  0  j 

   90  X C   90   X C j
 C   C 
Impedance

Because impedance is a complex quantity, it can be
represented graphically in the complex plane.
ZR = R0º = R + j0 = R
ZL = XL90º = 0 + jXL = jXL
ZC = XC-90º = 0 - jXC = -jXC
ELI the ICE man
E leads I
When voltage is applied to an
inductor, it resists the change
of current. The current builds
up more slowly, lagging in time
and phase.
I leads E
Since the voltage on a
capacitor is directly
proportional to the charge on it,
the current must lead the
voltage in time and phase to
conduct charge to the capacitor
plate and raise the voltage
Solving complex EE problems
1.
2.
3.
4.
Convert sine waves to phasors
Perform multiplication/division if needed
Convert phasors to complex numbers if
needed to perform addition/subtraction
Convert back to phasor form for the
answer to the problem
Important Notes



Peak values are useful for time domain
representations of signals.
RMS values are the standard when dealing
with phasor domain representations
If you need to represent something in the time
domain, you will need to convert RMS->Peak
voltage to obtain Em
E  VRMS 
 Em  VRMS 2
 e(t )  Em sin(2 ft   )
Example Problem 1
For the circuit below:
a) Time Domain voltage and current v(t) and i(t)
b) Draw the sine waveforms for v and i
c)
Draw the phasor diagram showing the relationship
between V and I
Example Problem 1
i(t)=50 sin (20000t) mA
v(t)=25 sin (20000t -90°) V
314 μsec
AC Series Circuits


Total impedance is sum of
individual impedances.
Also note that current is the
same through each element.
ZT  Z1  Z 2 
 Zn
Impedance example
Impedance example
ZT  ??????
ZT  3  4 j  5 j  3  j 
 3.2  18.4
Special case of Impedance

Whenever a capacitor and inductor of
equal reactances are placed in series, the
equivalent circuit is a short circuit.
Impedance

If the total impedance has only real
component, the circuit is said to be
resistive (X = 0 or  = 0°).
E  IZ
But since
 If  > 0°, the circuit is inductive. ELI
 If  < 0°, the circuit is capacitive. ICE

Example Problem 2
A network has a total impedance of ZT=24.0kΩ-30˚ at a
frequency of 2 kHz.
If the network consists of two series elements, what types
of components are those and what are their R/L/C
values?
Kirchhoff’s Voltage Law (KVL)

The phasor sum of voltage drops and rise
around a closed loop is equal to zero.
KVL
Vc  ??????
Vc  E  VR  VL
 100  8  36.87  1253.13
 6  126.87
Voltage Divider Rule (VDR)
Vx 
Zx
E
ZT
Vc  ??????
VDR


ZC
Vc  E 

 Z R  Z L  ZC 


3 j
 100 

 46j 3j 
 6  126.87
Example Problem 3
es(t)=170 sin (1000t + 0) V.

Determine ZTOT

Determine total current ITOT

Determine voltages VR, VL, and Vc

Verify KVL for this circuit

Graph E, VL, VC, VR in the time domain
Example Problem 3
180° phase
difference
between L
and C
Example Problem 4
es(t)=294 sin (377t + 0) V.

Use the voltage divider rule to find VL

Determine the value of inductance